I have been going through Pinter's A Book of Abstract Algebra recently and one question bugs me more than any other.
When discussing the properties of permutations on a general set $A$, he asks
Let $A$ be a finite set, and $B$ a subset of $A$. Let $G$ be the subset of $S_A$ (the symmetric group on $A$) consisting of all the permutations $f$ on $A$ such that $f(x)\in B$ for all $b\in B$. Prove that G is a subgroup of $S_A$.
I am fine with this, and I prove it this way:
i) if $f,g\in G$, then $(f\circ g)(x)=f(g(x))$, but $g(x)$ takes every $b\in B$ to some element in $B$. Similarly, if $f(x)$ takes in only the output of $g$, it will map all $b\in B$ to some element of $B$. Thus, $G$ is closed under composition of functions.
ii) Clearly, if $f$ takes any $b\in B$ to some element of $B$, then the same must be true for the permutations that "undoes" what $f$ did. Therefore, $f^{-1}$ is also in $G$.
From this, we can conclude $G$ is a subgroup of $S_A$.
The thing I do not understand is how this conclusion changes if $A$ is an infinite set. Pinter says there exists a counterexample to this, but I cannot find one. I.e., if $A$ an infinite set, it is not always true that $G$ is a subgroup of $S_A$.
Also, is my above proof valid; am I missing anything? It does not seem like I am using the assumption that $A$ is a finite set, so something tells me there must be a mistake somewhere.
Thank you in advance for any response.
edit: wording
Best Answer
Suppose $A$ is the set of integers and $B$ the set of positive integers. Then the shift function $f$ that maps $n$ to $n+1$ maps $B$ to itself but its inverse doesn't.
Now go back to your proof and see where you should use the finiteness of $A$.